On his 13th birthday, Josh’s aunt put some money in an account for him. The account pays 4% simple interest annually. She said,

It’s challenging to demonstrate that on here unless I sketch a diagram for you... It’s important to recognize that 10 thousandths equates to 1 hundredth. If you draw a square measuring 100 by 100 centimeters on graph paper, it will include a thousand individual squares because 100 x 100 equals 1000, and if you shade 10 of those squares, that represents 10 thousandths.

Hi! The goal of the Chi-Square Goodness of Fit test is to determine if observed frequencies of a categorical variable align with the expected historical or theoretical values in the population. Having the sales proportions of the top-five compact cars, we compare them against 400 compact car sales data from Chicago to see if there are discrepancies. Specifically, we have: - Chevy Cruze 24% ⟹ P(CC) = 0.24 - Ford Focus 21% ⟹ P(FF) = 0.21 - Hyundai Elantra 20% ⟹ P(HE) = 0.20 - Honda Civic 18% ⟹ P(HC) = 0.18 - Toyota Corolla 17% ⟹ P(TC) = 0.17 The hypotheses established are: H₀: P(CC) = 0.24; P(FF) = 0.21; P(HE) = 0.20; P(HC) = 0.18; P(TC) = 0.17 H₁: There is a discrepancy between expected and observed outcomes. With α set at 0.05, the statistic calculated is based on Oi (observed frequency) and Ei (expected frequency). The initial step involves calculating expected frequencies using: Ei = n * Pi, where Pi is the theoretical proportion for each category stated in the null hypothesis. The test conducted is right-tailed, and so is the p-value, calculated as: P(X²₄ ≥ 11.23) = 1 - P(X²₄ < 11.23) = 1 - 0.98 = 0.02. Since the p-value is lower than α, we reject the null hypothesis, indicating that Chicago's market shares for the five compact cars differ from those reported by Motor Trend.

Answer:

Step-by-step explanation: your friend would receive $17

and you would collect $21.25

I hope this helps!;))))

Response:

Detailed explanation:

Greetings!

You have the variable

X: Area eligible for painting with a can of spray paint (feet²)

This variable is normally distributed with a mean of μ= 25 feet² and a standard deviation of δ= 3 feet²

As this variable has a normal distribution, it needs to be converted into the standard normal form to utilize tabulated cumulative probabilities.

a.

P(X>27)

The first step involves standardizing the X value using Z= (X-μ)/ δ ~N(0;1)

P(Z>(27-25)/3)

P(Z>0.67)

Having determined the Z value, you can find it in the table, but since the table includes probabilities for , the following conversion must be applied:

P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143

b.

A sample of 20 cans was taken, and you need to ascertain the probability of averaging a coverage area of 540 feet².

The sample mean maintains the same distribution as its source variable, but its variance is influenced by sample size, thus it is normally distributed with parameters:

X[bar]~N(μ;δ²/n)

To cover 540 feet² with 20 cans, the average coverage must be approximately 540/20= 27 feet² per can.

c.

P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999

d.

No, if the distribution is not normal and skewed, the normal distribution should not be applied for calculating probabilities. While the central limit theorem might approximate the sampling distribution to normal when the sample size is 30 or larger, that isn’t applicable here.

I trust this information is helpful!

The volume provided is 3Pi(x^3) with a radius of x. To determine the volume of a cone, the formula used is V= [1/3]Pi(r^2)*height. By substituting, we get [1/3]Pi(r^2)x = 3Pi(x^3). This simplifies to (r^2)x = 9(x^3). Eventually, we find that r^2 = 9x^2, which leads to r = sqrt[9x^2] = 3x. <span>Answer: r = 3x</span>